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1 7.3 The Jacobi and Gauss-Seidel Iterative Methods

7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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Page 1: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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7.3 The Jacobi and Gauss-Seidel Iterative Methods

Page 2: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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The Jacobi Method Two assumptions made on Jacobi Method:

1.The system given by 𝑎𝑎11𝑥𝑥1 + 𝑎𝑎12𝑥𝑥2 + ⋯𝑎𝑎1𝑛𝑛𝑥𝑥𝑛𝑛 = 𝑏𝑏1 𝑎𝑎21𝑥𝑥1 + 𝑎𝑎22𝑥𝑥2 + ⋯𝑎𝑎2𝑛𝑛𝑥𝑥𝑛𝑛 = 𝑏𝑏2

⋮ 𝑎𝑎𝑛𝑛1𝑥𝑥1 + 𝑎𝑎𝑛𝑛2𝑥𝑥2 + ⋯𝑎𝑎𝑛𝑛𝑛𝑛𝑥𝑥𝑛𝑛 = 𝑏𝑏𝑛𝑛

Has a unique solution. 2.The coefficient matrix 𝐴𝐴 has no zeros on its main diagonal, namely, 𝑎𝑎11, 𝑎𝑎22, … , 𝑎𝑎𝑛𝑛𝑛𝑛 are nonzeros.

Page 3: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥1, the 2nd equation for 𝑥𝑥2 and so on to obtain the rewritten equations:

𝑥𝑥1 =1𝑎𝑎11

(𝑏𝑏1 − 𝑎𝑎12𝑥𝑥2 − 𝑎𝑎13𝑥𝑥3 − ⋯𝑎𝑎1𝑛𝑛𝑥𝑥𝑛𝑛)

𝑥𝑥2 =1𝑎𝑎22

(𝑏𝑏2 − 𝑎𝑎21𝑥𝑥1 − 𝑎𝑎23𝑥𝑥3 − ⋯𝑎𝑎2𝑛𝑛𝑥𝑥𝑛𝑛)

𝑥𝑥𝑛𝑛 =1𝑎𝑎𝑛𝑛𝑛𝑛

(𝑏𝑏𝑛𝑛 − 𝑎𝑎𝑛𝑛1𝑥𝑥1 − 𝑎𝑎𝑛𝑛2𝑥𝑥2 − ⋯𝑎𝑎𝑛𝑛,𝑛𝑛−1𝑥𝑥𝑛𝑛−1)

Then make an initial guess of the solution 𝒙𝒙(0) = (𝑥𝑥1(0), 𝑥𝑥2

(0), 𝑥𝑥3(0), … 𝑥𝑥𝑛𝑛

(0)). Substitute these values into the right hand side the of the rewritten equations to obtain the first approximation, �𝑥𝑥1

(1), 𝑥𝑥2(1), 𝑥𝑥3

(1), … 𝑥𝑥𝑛𝑛(1)�.

This accomplishes one iteration.

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In the same way, the second approximation �𝑥𝑥1(2), 𝑥𝑥2

(2), 𝑥𝑥3(2), … 𝑥𝑥𝑛𝑛

(2)� is computed by substituting the first approximation’s value �𝑥𝑥1

(1), 𝑥𝑥2(1), 𝑥𝑥3

(1), … 𝑥𝑥𝑛𝑛(1)� into the right hand side of the rewritten equations.

By repeated iterations, we form a sequence of approximations 𝒙𝒙(𝑘𝑘) =

�𝑥𝑥1(𝑘𝑘), 𝑥𝑥2

(𝑘𝑘), 𝑥𝑥3(𝑘𝑘), … 𝑥𝑥𝑛𝑛

(𝑘𝑘)�𝑡𝑡

, 𝑘𝑘 = 1,2,3, …

The Jacobi Method. For each 𝑘𝑘 ≥ 1, generate the components 𝑥𝑥𝑖𝑖(𝑘𝑘) of

𝒙𝒙(𝒌𝒌) from 𝒙𝒙(𝒌𝒌−𝟏𝟏) by

𝑥𝑥𝑖𝑖(𝑘𝑘) =

1𝑎𝑎𝑖𝑖𝑖𝑖

⎣⎢⎢⎢⎡�(−𝑎𝑎𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖

(𝑘𝑘−1))𝑛𝑛

𝑖𝑖=1,𝑖𝑖≠𝑖𝑖

+ 𝑏𝑏𝑖𝑖

⎦⎥⎥⎥⎤

, for 𝑖𝑖 = 1,2, , …𝑛𝑛

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Example. Apply the Jacobi method to solve 5𝑥𝑥1 − 2𝑥𝑥2 + 3𝑥𝑥3 = −1−3𝑥𝑥1 + 9𝑥𝑥2 + 𝑥𝑥3 = 2 2𝑥𝑥1 − 𝑥𝑥2 − 7𝑥𝑥3 = 3

Continue iterations until two successive approximations are identical when rounded to three significant digits. Solution

𝑛𝑛 𝑘𝑘 = 0 𝑘𝑘 = 1 𝑘𝑘 = 2 𝑘𝑘 = 3 𝑘𝑘 = 4 𝑘𝑘 = 5 𝑘𝑘 = 6 𝑥𝑥1

(𝑘𝑘) 0.000 -0.200 0.146 0.192

𝑥𝑥2(𝑘𝑘) 0.000 0.222 0.203 0.328

𝑥𝑥2(𝑘𝑘) 0.000 -0.429 -0.517 -0.416

Page 6: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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When to stop: 1. ||𝒙𝒙(𝑘𝑘)−𝒙𝒙(𝑘𝑘−1)||

||𝒙𝒙(𝑘𝑘)||< 𝜀𝜀; or ��𝒙𝒙(𝑘𝑘) − 𝒙𝒙(𝑘𝑘−1)�� < 𝜀𝜀. Here 𝜀𝜀 is

a given small number . Another stopping criterion: ||𝒙𝒙(𝑘𝑘)−𝒙𝒙(𝑘𝑘−1)||

||𝒙𝒙(𝑘𝑘)||

Definition 7.1 A vector norm on 𝑅𝑅𝑛𝑛 is a function, || ∙ ||, from 𝑅𝑅𝑛𝑛 to 𝑅𝑅 with the properties:

(i) ||𝒙𝒙|| ≥ 0 for all 𝒙𝒙 ∈ 𝑅𝑅𝑛𝑛 (ii) �|𝒙𝒙|� = 0 if and only if 𝒙𝒙 = 𝟎𝟎 (iii)�|𝛼𝛼𝒙𝒙|� = |𝛼𝛼|�|𝒙𝒙|� for all 𝛼𝛼 ∈ 𝑅𝑅 and 𝒙𝒙 ∈ 𝑅𝑅𝑛𝑛 (iv) �|𝒙𝒙 + 𝒚𝒚|� ≤ �|𝒙𝒙|� + ||𝒚𝒚|| for all 𝒙𝒙,𝒚𝒚 ∈ 𝑅𝑅𝑛𝑛

Definition 7.2 The Euclidean norm 𝑙𝑙2 and the infinity norm 𝑙𝑙∞ for the vector 𝒙𝒙 = [𝑥𝑥1, 𝑥𝑥2, … , 𝑥𝑥𝑛𝑛]𝑡𝑡 are defined by

||𝒙𝒙||2 = ��𝑥𝑥𝑖𝑖2𝑛𝑛

𝑖𝑖=1

12

and

Page 7: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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||𝒙𝒙||∞ = max1≤𝑖𝑖≤𝑛𝑛

|𝑥𝑥𝑖𝑖|

Example. Determine the 𝑙𝑙2 and 𝑙𝑙∞ norms of the vector 𝑥𝑥 = (−1,1,−2)𝑡𝑡. Solution:

||𝒙𝒙||2 = �(−1)2 + (1)2 + (−2)2 = √6.

||𝒙𝒙||∞ = max{|−1|, |1|, |−2|} = 2.

Page 8: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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The Jacobi Method in Matrix Form Consider to solve an 𝑛𝑛 × 𝑛𝑛 size system of linear equations 𝐴𝐴𝒙𝒙 = 𝒃𝒃 with

𝐴𝐴 = �

𝑎𝑎11 𝑎𝑎12 … 𝑎𝑎1𝑛𝑛𝑎𝑎21 𝑎𝑎22 … 𝑎𝑎2𝑛𝑛⋮𝑎𝑎𝑛𝑛1

⋮𝑎𝑎𝑛𝑛2

⋱…

⋮𝑎𝑎𝑛𝑛𝑛𝑛

� and 𝒃𝒃 = �

𝑏𝑏1𝑏𝑏2⋮𝑏𝑏𝑛𝑛

� for 𝒙𝒙 = �

𝑥𝑥1𝑥𝑥2⋮𝑥𝑥𝑛𝑛�.

We split 𝐴𝐴 into

Page 9: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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𝐴𝐴 = �

𝑎𝑎11 0 … 00 𝑎𝑎22 … 0⋮0

⋮0

⋱…

⋮𝑎𝑎𝑛𝑛𝑛𝑛

� − �

0 … 0 0−𝑎𝑎21 … 0 0⋮

−𝑎𝑎𝑛𝑛1⋱ ⋮

… −𝑎𝑎𝑛𝑛,𝑛𝑛−1 0

− �

0 −𝑎𝑎12 … −𝑎𝑎1𝑛𝑛0 0 ⋮⋮0

⋮0

⋱…

−𝑎𝑎𝑛𝑛−1,𝑛𝑛0

� = 𝐷𝐷 − 𝐿𝐿 − 𝑈𝑈

Where 𝐷𝐷 = �

𝑎𝑎11 0 … 00 𝑎𝑎22 … 0⋮0

⋮0

⋱…

⋮𝑎𝑎𝑛𝑛𝑛𝑛

� 𝐿𝐿 = �

0 … 0 0−𝑎𝑎21 … 0 0⋮

−𝑎𝑎𝑛𝑛1⋱ ⋮

… −𝑎𝑎𝑛𝑛,𝑛𝑛−1 0

�,

Page 10: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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𝑈𝑈 = �

0 −𝑎𝑎12 … −𝑎𝑎1𝑛𝑛0 0 ⋮⋮0

⋮0

⋱…

−𝑎𝑎𝑛𝑛−1,𝑛𝑛0

𝐴𝐴𝒙𝒙 = 𝒃𝒃 is transformed into (𝐷𝐷 − 𝐿𝐿 − 𝑈𝑈)𝒙𝒙 = 𝒃𝒃. 𝐷𝐷𝒙𝒙 = (𝐿𝐿 + 𝑈𝑈)𝒙𝒙 + 𝒃𝒃

Assume 𝐷𝐷−1 exists and 𝐷𝐷−1 =

⎣⎢⎢⎢⎢⎡1𝑎𝑎11

0 … 0

0 1𝑎𝑎22

… 0

⋮0

⋮0

⋱…

⋮1𝑎𝑎𝑛𝑛𝑛𝑛⎦

⎥⎥⎥⎥⎤

Then 𝒙𝒙 = 𝐷𝐷−1(𝐿𝐿 + 𝑈𝑈)𝒙𝒙 + 𝐷𝐷−1𝒃𝒃

The matrix form of Jacobi iterative method is 𝒙𝒙(𝑘𝑘) = 𝐷𝐷−1(𝐿𝐿 + 𝑈𝑈)𝒙𝒙(𝑘𝑘−1) + 𝐷𝐷−1𝒃𝒃 𝑘𝑘 = 1,2,3, …

Page 11: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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Define 𝑇𝑇𝑖𝑖 = 𝐷𝐷−1(𝐿𝐿 + 𝑈𝑈) and 𝒄𝒄 = 𝐷𝐷−1𝒃𝒃, Jacobi iteration method can also be written as

𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝑖𝑖𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄 𝑘𝑘 = 1,2,3, … The Gauss-Seidel Method Main idea of Gauss-Seidel

Page 12: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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With the Jacobi method, only the values of 𝑥𝑥𝑖𝑖(𝑘𝑘) obtained in the 𝑘𝑘 th

iteration are used to compute 𝑥𝑥𝑖𝑖(𝑘𝑘+1). With the Gauss-Seidel method, we

use the new values 𝑥𝑥𝑖𝑖(𝑘𝑘+1) as soon as they are known. For example, once

we have computed 𝑥𝑥1(𝑘𝑘+1) from the first equation, its value is then used in

the second equation to obtain the new 𝑥𝑥2(𝑘𝑘+1), and so on.

Example. Use the Gauss-Seidel method to solve

Page 13: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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5𝑥𝑥1 − 2𝑥𝑥2 + 3𝑥𝑥3 = −1−3𝑥𝑥1 + 9𝑥𝑥2 + 𝑥𝑥3 = 2 2𝑥𝑥1 − 𝑥𝑥2 − 7𝑥𝑥3 = 3

Choose the initial guess 𝑥𝑥1 = 0, 𝑥𝑥2 = 0, 𝑥𝑥3 = 0 𝑛𝑛 𝑘𝑘 = 0 𝑘𝑘 = 1 𝑘𝑘 = 2 𝑘𝑘 = 3 𝑘𝑘 = 4 𝑘𝑘 = 5 𝑘𝑘 = 6 𝑥𝑥1

(𝑘𝑘) 0.000 -0.200 0.167

𝑥𝑥2(𝑘𝑘) 0.000 0.156 0.334

𝑥𝑥2(𝑘𝑘) 0.000 -0.508 -0.429

The Gauss-Seidel Method. For each 𝑘𝑘 ≥ 1, generate the components 𝑥𝑥𝑖𝑖

(𝑘𝑘) of 𝒙𝒙(𝒌𝒌) from 𝒙𝒙(𝒌𝒌−𝟏𝟏) by

Page 14: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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𝑥𝑥𝑖𝑖(𝑘𝑘) =

1𝑎𝑎𝑖𝑖𝑖𝑖

�−�(𝑎𝑎𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖(𝑘𝑘))

𝑖𝑖−1

𝑖𝑖=1

− � (𝑎𝑎𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖(𝑘𝑘−1))

𝑛𝑛

𝑖𝑖=𝑖𝑖+1

+ 𝑏𝑏𝑖𝑖� , for 𝑖𝑖

= 1,2, , …𝑛𝑛 Namely,

𝑎𝑎11𝑥𝑥1(𝑘𝑘) = −𝑎𝑎12𝑥𝑥2

(𝑘𝑘−1) −⋯− 𝑎𝑎1𝑛𝑛𝑥𝑥𝑛𝑛(𝑘𝑘−1) + 𝑏𝑏1

𝑎𝑎22𝑥𝑥2(𝑘𝑘) = −𝑎𝑎21𝑥𝑥1

(𝑘𝑘) − 𝑎𝑎23𝑥𝑥3(𝑘𝑘−1) −⋯− 𝑎𝑎2𝑛𝑛𝑥𝑥𝑛𝑛

(𝑘𝑘−1) + 𝑏𝑏2𝑎𝑎33𝑥𝑥3

(𝑘𝑘) = −𝑎𝑎31𝑥𝑥1(𝑘𝑘) − 𝑎𝑎32𝑥𝑥2

(𝑘𝑘) − 𝑎𝑎34𝑥𝑥4(𝑘𝑘−1) −⋯− 𝑎𝑎3𝑛𝑛𝑥𝑥𝑛𝑛

(𝑘𝑘−1) + 𝑏𝑏3⋮

𝑎𝑎𝑛𝑛𝑛𝑛𝑥𝑥𝑛𝑛(𝑘𝑘) = −𝑎𝑎𝑛𝑛1𝑥𝑥1

(𝑘𝑘) − 𝑎𝑎𝑛𝑛2𝑥𝑥2(𝑘𝑘) −⋯− 𝑎𝑎𝑛𝑛,𝑛𝑛−1𝑥𝑥𝑛𝑛−1

(𝑘𝑘) + 𝑏𝑏𝑛𝑛

Matrix form of Gauss-Seidel method.

Page 15: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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(𝐷𝐷 − 𝐿𝐿)𝒙𝒙(𝒌𝒌) = 𝑈𝑈𝒙𝒙(𝑘𝑘−1) + 𝒃𝒃 𝒙𝒙(𝒌𝒌) = (𝐷𝐷 − 𝐿𝐿)−1𝑈𝑈𝒙𝒙(𝑘𝑘−1) + (𝐷𝐷 − 𝐿𝐿)−1𝒃𝒃

Define 𝑇𝑇𝑔𝑔 = (𝐷𝐷 − 𝐿𝐿)−1𝑈𝑈 and 𝒄𝒄𝑔𝑔 = (𝐷𝐷 − 𝐿𝐿)−1𝒃𝒃, Gauss-Seidel method can be written as

𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝑔𝑔𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄𝑔𝑔 𝑘𝑘 = 1,2,3, … Convergence theorems of the iteration methods

Page 16: 7.3 The Jacobi and Gauss-Seidel Iterative Methodszxu2/acms40390F15/Lec-7.3.pdf · Main idea of Jacobi To begin, solve the 1st equation for 𝑥𝑥 1, the 2 nd equation for 𝑥𝑥

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Let the iteration method be written as 𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄 for each 𝑘𝑘 = 1,2,3, …

Definition 7.14 The spectral radius 𝜌𝜌(𝐴𝐴) of a matrix 𝐴𝐴 is defined by

𝜌𝜌(𝐴𝐴) = max|𝜆𝜆| , where 𝜆𝜆 is an eigenvalue of 𝐴𝐴. Remark: For complex 𝜆𝜆 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏, we define |𝜆𝜆| = √𝑎𝑎2 + 𝑏𝑏2. Lemma 7.18 If the spectral radius satisfies 𝜌𝜌(𝑇𝑇) < 1, then (𝐼𝐼 − 𝑇𝑇)−1 exists, and

(𝐼𝐼 − 𝑇𝑇)−1 = 𝐼𝐼 + 𝑇𝑇 + 𝑇𝑇2 + ⋯ = �𝑇𝑇𝑖𝑖∞

𝑖𝑖=0

Theorem 7.19 For any 𝒙𝒙(0) ∈ 𝑅𝑅𝑛𝑛, the sequence {𝒙𝒙(𝑘𝑘)}𝑘𝑘=0∞ defined by

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𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄 for each 𝑘𝑘 ≥ 1 converges to the unique solution of 𝒙𝒙 = 𝑇𝑇𝒙𝒙 + 𝒄𝒄 if and only if 𝜌𝜌(𝑇𝑇) < 1. Proof (only show 𝜌𝜌(𝑇𝑇) < 1 is sufficient condition) 𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄 = 𝑇𝑇�𝑇𝑇𝒙𝒙(𝑘𝑘−2) + 𝒄𝒄� + 𝒄𝒄 = ⋯ = 𝑇𝑇𝑘𝑘𝒙𝒙(0) + (𝑇𝑇𝑘𝑘−1 +⋯+ 𝑇𝑇 + 𝐼𝐼)𝒄𝒄 Since 𝜌𝜌(𝑇𝑇) < 1, lim

𝑘𝑘→∞𝑇𝑇𝑘𝑘𝒙𝒙(0) = 𝟎𝟎

lim𝑘𝑘→∞

𝒙𝒙(𝑘𝑘) = 𝟎𝟎 + lim𝑘𝑘→∞

(�𝑇𝑇𝑖𝑖𝑘𝑘−1

𝑖𝑖=0

) 𝒄𝒄 = (𝐼𝐼 − 𝑇𝑇)−1𝒄𝒄

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Definition 7.8 A matrix norm || ∙ || on 𝑛𝑛 × 𝑛𝑛 matrices is a real-valued function satisfying

(i) ||𝐴𝐴|| ≥ 0 (ii) �|𝐴𝐴|� = 0 if and only if 𝐴𝐴 = 0 (iii)�|𝛼𝛼𝐴𝐴|� = |𝛼𝛼|�|𝐴𝐴|� (iv) �|𝐴𝐴 + 𝐵𝐵|� ≤ �|𝐴𝐴|� + ||𝐵𝐵|| (v) �|𝐴𝐴𝐵𝐵|� ≤ �|𝐴𝐴|�||𝐵𝐵||

Theorem 7.9. If || ∙ || is a vector norm, the induced (or natural) matrix norm is given by

�|𝐴𝐴|� = max�|𝒙𝒙|�=1

||𝐴𝐴𝒙𝒙||

Example. ||𝐴𝐴||∞ = max

||𝒙𝒙||∞=1||𝐴𝐴𝒙𝒙||∞, the 𝑙𝑙∞ induced norm.

||𝐴𝐴||2 = max||𝒙𝒙||2=1

||𝐴𝐴𝒙𝒙||2, the 𝑙𝑙2 induced norm.

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Theorem 7.11. If 𝐴𝐴 = [𝑎𝑎𝑖𝑖𝑖𝑖] is an 𝑛𝑛 × 𝑛𝑛 matrix, then

�|𝐴𝐴|�∞ = max1≤𝑖𝑖≤𝑛𝑛

��𝑎𝑎𝑖𝑖𝑖𝑖�𝑛𝑛

𝑖𝑖=1

Example. Determine �|𝐴𝐴|�∞ for the matrix 𝐴𝐴 = �1 2 −10 3 −15 −1 1

Corollary 7.20 If �|𝑇𝑇|� < 1 for any natural matrix norm and 𝒄𝒄 is a given vector, then the sequence {𝒙𝒙(𝑘𝑘)}𝑘𝑘=0∞ defined by 𝒙𝒙(𝑘𝑘) = 𝑇𝑇𝒙𝒙(𝑘𝑘−1) + 𝒄𝒄 converges, for any 𝒙𝒙(0) ∈ 𝑅𝑅𝑛𝑛 , to a vector 𝒙𝒙 ∈ 𝑅𝑅𝑛𝑛, with 𝒙𝒙 = 𝑇𝑇𝒙𝒙 + 𝒄𝒄 , and the following error bound hold:

(i) |�𝒙𝒙 − 𝒙𝒙(𝑘𝑘)�| ≤ �|𝑇𝑇|�𝑘𝑘|�𝒙𝒙(0) − 𝒙𝒙�|

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(ii) |�𝒙𝒙 − 𝒙𝒙(𝑘𝑘)�| ≤ �|𝑇𝑇|�𝑘𝑘

1−||𝑇𝑇|||�𝒙𝒙(1) − 𝒙𝒙(0)�|

Theorem 7.21 If 𝐴𝐴 is strictly diagonally dominant, then for any choice of 𝒙𝒙(0), both the Jacobi and Gauss-Seidel methods give sequences {𝒙𝒙(𝑘𝑘)}𝑘𝑘=0∞ that converges to the unique solution of 𝐴𝐴𝒙𝒙 = 𝒃𝒃. Rate of Convergence Corollary 7.20 (i) implies |�𝒙𝒙 − 𝒙𝒙(𝑘𝑘)�| ≈ 𝜌𝜌(𝑇𝑇)𝑘𝑘|�𝒙𝒙(0) − 𝒙𝒙�| Theorem 7.22 (Stein-Rosenberg) If 𝑎𝑎𝑖𝑖𝑖𝑖 ≤ 0, for each 𝑖𝑖 ≠ 𝑏𝑏 and 𝑎𝑎𝑖𝑖𝑖𝑖 ≥ 0, for each 𝑖𝑖 = 1,2, … ,𝑛𝑛 , then one and only one of following statements holds:

(i) 0 ≤ 𝜌𝜌�𝑇𝑇𝑔𝑔� < 𝜌𝜌�𝑇𝑇𝑖𝑖� < 1; (ii) 1 < 𝜌𝜌�𝑇𝑇𝑖𝑖� < 𝜌𝜌�𝑇𝑇𝑔𝑔�; (iii)𝜌𝜌�𝑇𝑇𝑖𝑖� = 𝜌𝜌�𝑇𝑇𝑔𝑔� = 0; (iv) 𝜌𝜌�𝑇𝑇𝑖𝑖� = 𝜌𝜌�𝑇𝑇𝑔𝑔� = 1.

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